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$\newcommand{\R}{\mathbb{R}}$ I am trying to follow a derivation of the first variation formula for the energy functional. (In "Selected Topics in Harmonic maps").

Here is the context:

$M,N$ are Riemannian manifolds, $\phi:M \to N$ is a smooth map which is a critical point of the energy functional $\int_M |d\phi|^2$ where $|\cdot|$ is the natural induced norm (inner product) on $T^*M \otimes \phi^*(TN)$.($\phi^*(TN)$ is the pullback bundle of $TN$ along $\phi$)

The author then defines a family of maps $\phi_t:M \to N$ ($\phi_0=\phi$),

i.e a map $\psi:M \times \R \to N \, , \, \psi(p,t)=\phi_t(p)$

$$0=\frac{1}{2}\frac{d}{dt}\int_M |d\phi_t|^2 \text{vol}_g = \frac{1}{2}\int_M \frac{\partial}{\partial t}\langle d\phi_t,d\phi_t \rangle \text{vol}_g \stackrel{(*)}{=} \int_M \langle \nabla_{\frac{\partial}{\partial t}} d\phi_t,d\phi_t \rangle \text{vol}_g $$

(where all the derivatives are taken at time $t=0$)

Here is my problem in a nutshell:

The metric and connection in the two sides of $(*)$ are not defined on the same bundles.

Details:

The author mentions that the connection $\nabla$ in the formula above, is the induced one on $T^*(M \times \R)\otimes \psi^*(TN)$

I guess equality $(*)$ follows from metricity (All the original connections are Levi-Civita, hence they and everything induced by them is metric).

But this metricity is w.r.t the metric on $T^*(M \times \R)\otimes \psi^*(TN)$, hence instead of $\langle d\phi_t,d\phi_t \rangle$ I think it's supposed to be written $\langle d\psi,d\psi \rangle$, where this is the (squared) norm of the differential of the map $\psi$, i.e at each point $(p,t)$ this is the induced norm on $\operatorname{Hom}(T_{(p,t)}(M \times \R),T_{\phi_t(p)}N)$

(This is the only way I am able to make sense of this. I am assuming that $\R$ is endowed with its standard metric, and $M \times \R$ with the product metric).

The problem is that now there is no reason this should be zero, since the variational assumption is that $\frac{d}{dt}\int_M |d\phi_t|^2 \text{vol}_g=0$ , not $\frac{d}{dt}\int_M |d\psi|^2 \text{vol}_g=0$, and I see no apparent reason why it should hold that $|d\psi|=|d\phi_t|$ in general.

Am I missing something? what is going on here?

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If $E \to X$ is a vector bundle over a manifold $X$ with metric $\langle -,-\rangle$ and connection $\nabla$, metric compatibility is the condition $$ \xi \langle s_1, s_2 \rangle = \langle \nabla_\xi s_1, s_2 \rangle + \langle s_1, \nabla_\xi s_2 \rangle $$ for any vector field $\xi$ on $X$ and any two sections $s_1, s_2$ of $E$.

In your case, we can think of $d\phi$ (with $t$ varying) as a section of the bundle $p^\ast(T^\ast M) \otimes \psi^\ast(TN)$ over $M \times \mathbb{R}$, where $p: M \times \mathbb{R} \to M$ is the projection onto the first factor. This bundle has a natural metric and connection induced by those on $TM$ and $TN$. This connection is compatible with the metric, so for any vector field $\xi$ we have $$ \xi |d\phi|^2 = 2 \langle \nabla_\xi d\phi, d\phi \rangle $$ Take $\xi = \partial / \partial t$ to obtain (*).

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